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        riddles >> hard >> Binary Implications and Binomial Parity
        
(Message started by: william wu on Oct 29^th, 2003, 7:35pm)

Title: Binary Implications and Binomial Parity
Post by william wu on Oct 29^th, 2003, 7:35pm

Let r and s be natural numbers. Denote the binary expansion of r by r₁r₂...r_n. Similarly, denote the binary expansion of s by s₁s₂...s_n. Assume these binary expansions are both of length n.

Let the relation r [smiley=rrightarrow.gif] s mean that the following sentence is true:

[forall]i[in]{1,...,n}, IF r_i = 1, THEN s_i = 1

Thus for example, for binary strings 1010 and 1110, it follows that 1010 [smiley=rrightarrow.gif] 1110.

Prove that

r [smiley=rrightarrow.gif] s IF AND ONLY IF C(s,r) is odd

where C(s,r) is the binomial coefficient: (s!)/(r!(s-r)!)

Note 1: An amazing result from the 1900s! It was passed to me via a colleague (Jon Yard). Does anyone know how it is proven (I don't), or what the theorem's name is if it has a name? If anyone knows the proof and feels it is too hard for recreational math, please say so.

Note 2: This result apparently has a little to do with Hilbert's 10th problem.

Title: Re: Binary Implications and Binomial Parity
Post by Rezyk on Oct 29^th, 2003, 10:44pm

Tool #1: [hide]The number of times 2 divides X! is equal to X minus the number of 1s in the binary representation of X.[/hide] (proof omitted)

Tool #2: [hide]When adding 2 numbers in binary, iff there is no carry anywhere, the number of "1" digits is conserved.[/hide] (proof omitted)

C(s, r) is odd
[bigleftrightarrow] [hide]s!/(r!(s-r)!) is not divisible by 2[/hide]
[bigleftrightarrow] [hide]number of times 2 divides s! = (number of times 2 divides r!)+(number of times 2 divides (s-r)!)[/hide]
[bigleftrightarrow] [hide]s - (number of 1s in binary repr. of s) = r - (number of 1s in binary repr. of r) + s-r - (number of 1s in binary repr. of s-r)[/hide]
[bigleftrightarrow] [hide]number of 1s in binary repr. of s = (number of 1s in binary repr. of r) + (number of 1s in binary repr. of s-r)[/hide]
[bigleftrightarrow] [hide]there is no carry anywhere in adding r to (s-r) in binary[/hide]
[bigleftrightarrow] r [smiley=rrightarrow.gif] s

Title: Re: Binary Implications and Binomial Parity
Post by Barukh on Oct 30^th, 2003, 11:05am

Awesome solution, Rezyk! And really elementary. Let me add something for the completeness of the derivation:

Tool #1 - proof:
[smiley=blacksquare.gif]
[hide]Denote F(n) the maximal power of 2 dividing n!, P(n) - number of 1's in the binary representation (BR) of n. We want to show that F(n) + P(n) = n. Use induction by n:

1. F(1) + P(1) = 1 + 0 = 1.
2. Let the rightmost 1 in BR of (n+1) be at position j. Then, the j-th position in BR of n was 0, but positions 1, ..., j-1 were all 1's. So, P(n+1) = P(n) - (j-2). Also, 2^j-1 divides n+1, so F(n+1) = F(n) + (j-1). Therefore, F(n+1) + P(n+1) = n+1.[/hide]
[smiley=blacksquare.gif]